[Numpy-discussion] from_function
Yakov Keselman
yakov.keselman at gmail.com
Tue Feb 10 16:11:36 EST 2009
Perhaps you can do something along the following lines to get around
this limitation:
#################
# parameterizes the original function by delta, size.
def parameterized_function(delta, size, function):
center = (size-1)/2.0
return lambda i: function( (i-center)*delta )
# the function to which we want to apply "fromfunction".
def square(x): return x*x
# test script
from numpy import fromfunction
Delta = 0.1
Size = 9
print fromfunction( parameterized_function(Delta, Size, square), (Size,) )
###########
[ 0.16 0.09 0.04 0.01 0. 0.01 0.04 0.09 0.16]
On 2/3/09, Neal Becker <ndbecker2 at gmail.com> wrote:
> I've been using something I wrote:
>
> coef_from_function (function, delta, size)
> which does (c++ code):
>
> double center = double(size-1)/2;
> for (int i = 0; i < size; ++i)
> coef[i] = call<value_t> (func, double(i - center) * delta);
>
> I thought to translate this to np.fromfunction. It seems fromfunction is
> not as flexible, it uses only a fixed integer grid?
>
>
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